Question

(a) Calculate the molarity of hydrogen chloride in a solution when $0.365{\text{ }}g$ of it has been dissolved in $100{\text{ }}mL$ of solution?
(b) $3.0{\text{ }}g$ of a salt of molecular weight $30$ is dissolved in $250{\text{ }}ml$ water. The molality of solution is:

Answer 1

Hint : Before going through the question let us first know about the key terms related to this question i.e. molarity and molality. The number of solvent moles present in $1{\text{ }}L$ of solution is defined as molarity. The volume of solutions comprises the denominator. The volume depends on the system temperature. The number of solute moles that are present in $1{\text{ }}kg$ of solvent is defined as molality. This unit is made up of mass conditions and is temperature independent.

Complete Step By Step Answer:
Now, let us solve the questions accordingly;
(a) The molarity of hydrogen chloride in a solution when $0.365{\text{ }}g$ of it has been dissolved in $100{\text{ }}mL$ of solution is:
Molarity $\left( M \right)$ = $\dfrac{n}{v}$
$M =$ Molar concentration
$n =$ Moles of solute
$v =$ Litres of solution
No. of moles $(n) =$ $\dfrac{{mass}}{{molar\,mass}}$
For $HCl$ ;
Mass = $0.365{\text{ }}g$
Molar mass = ${\text{36}}{\text{.5 g/mol }}$
Volume = $100mL = 0.1{\text{ }}L$
Thus, molarity = $\dfrac{{0.365}}{{36.5 \times 0.1}} = 0.1M$
Therefore, the molarity of hydrogen chloride is $0.1M$ .
(b) $3.0{\text{ }}g$ of a salt of molecular weight $30$ is dissolved in $250{\text{ }}ml$ water. The molality of solution is
The term for solution molality is:
= \dfrac{{Number\,of\,moles\,of\,solute}}{{Weight\,of\,solvent\,in\,Kg}} \\\ = \dfrac{{Weight\,of\,solute}}{{Molar\,mass}} \times \dfrac{{1000}}{{Weight\,of\,solvent\,in\,Kg}} \\\ = \dfrac{3}{{30}} \times \dfrac{{1000}}{{250}} = 0.4M \\\
Therefore, the molality of the solution is $0.4M$ .

Note :
Although molarity is usually used to express concentrations for solution reactions or for titrations, there is one drawback — molarity is the number of solute moles divided by solution volume, and the solution volume varies according to the density of the solution, depending on the temperature.